# Holomorphism of $f$ and $f^n$

Given $f^3$ and $f^7$ holomorphic functions, I want to show that $f$ is holomorphic.

Can I say that if $f$ is not holomorphic then $f^3$ is not holomorphic which mean $f$ is must be holomorphic?

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Is $f^n = f \cdot f \cdot \dots \cdot f$ or $f^n = f \circ f \circ \dots \circ f$? – Alexander Thumm Aug 16 '12 at 10:38
To your last bit, no: consider ${}^3\sqrt{x}$. It most often doesn't work to ignore half the given information. – Kevin Carlson Aug 16 '12 at 10:40
Even more drastic: take any function $f: \mathbb C \to \{ 1, \zeta, \zeta^2\}$, where $\zeta$ is a nontrivial third root of unity. – Alexander Thumm Aug 16 '12 at 10:44

As David pointed out, $f(z)=\frac{f^7(z)}{f^6(z)}$ is holomorphic at all points where $f^6(z)\neq0$. Using Riemann's theorem on removable singularities it's enough to show that $\lim_{z\to0}f(z)$ exists. Now, assume that $f^6(z_0)=0$. Since $f^6$ is holomorphic at $z_0$, you can write $f^6(z)=(z-z_0)^kg(z)$ where $g$ is holomorphic and $g(z_0)\neq0$. Also, $f^6(z_0)=0$ implies that $f^7(z_0)=0$, so you can write $f^7(z)=(z-z_0)^mh(z)$. Now consider $f^{42}$: it is holomorphic, as a product of holomorphic and hence $(z-z_0)^{7k}g(z)=(z-z_0)^{6m}h(z)$. Since the order of the zero at $z_0$ is constant, we have $7k=6m$, so $m>k$. Hence you can write $f(z)=(z-z_0)^{m-k}\frac{h(z)}{g(z)}$, and $g$ doesn't vanish at $z_0$, so the limit of $f(z)$ at $z_0$ exists (and is equal to $0$).